1. 2. Open and Closed SetsThroughout the following (M, d) will be taken to be a metric space. We introduce here thenotion of open balls and, in due course, open sets which will allow us to rephrase ourdefinition of continuity.Notation 25 Let x M and > 0. Then the open ball B(x, ) and closed ball B(x, ) aredefined asB(x, ) = {y M : d(x, y) < } , B(x, ) = {y M : d(x, y) } .If we wish to stress that we are working in M we will write BM(x, ) and BM(x, ).We can see that Definition 19, regarding the continuity of a function f : M N betweenmetric spaces can be rephrased as: f is continuous at x if > 0 > 0 f(BM(x, )) BN(f(x), ). (2.1)This rephrasing will become still more streamlined and elegant once we have introduced themore general idea of open sets.Definition 26 U M is said to be open in M (or simply open) ifx U > 0 B(x, ) U.We say that x is an interior point of U if there exists > 0 such that B(x, ) U, so sayingU is open is equivalent to saying all its points are interior points.So (2.1) can be rephrased as: f : M N is continuous at x ifwhenever f(x) is an interior point of U N then x is an interior point of f1(U) Mor can be rephrased still more cleanly when we are talking about a function which is continuouseverywhere on M as:Theorem 27 Let f : M N be a map between metric spaces (M, dM) and (N, dN). Then fis continuous if and only if whenever U N is open then f1(U) M is open.Remark 28 Hopefully this helps motivate the importance of open sets as, with this result, wesee they are sufficient to determine which functions are continuous.Proof. Suppose that f is continuous and that U is an open subset of N. Let x f1(U) sothat f(x) is an interior point of U (as U is open). So there exists > 0 such thatBN(f(x), ) U.OPEN AND CLOSED SETS 13

2. As f is continuous at x there exists > 0 such thatf(BM(x, )) BN(f(x), ) Uand hence BM(x, ) f1(U), showing that x is an interior point of f1(U). As x was anarbitrary point, then f1(U) is open.Conversely suppose that the pre-image of every open set in N is open in M. Let > 0 andx M. As BN(f(x), ) is open in N, by assumption x is an interior point of f1(BN(f(x), ))and so there exists > 0 such thatBM(x,) f1(BN(f(x), )) and hence f(BM(x,)) BN(f(x), ),which is just another way of saying the f is continuous at x.Definition 29 We refer to the open sets of M as the topology of M, or the topology inducedby the metric d.Proposition 30 (a) The intersection of finitely many open sets is open.(b) An arbitrary union of open sets is open.Proof. (a) Let U1, . . . ,Um be (finitely many) open subsets of M. Let x mi=1Ui. Then, foreach i, there exists i > 0 such that B(x, i) Ui. Let = min {1, 2, . . . , m} > 0 and thenB(x, ) mi=1Ui.(b) For i I (a not-necessarily finite indexing set), let Ui be an open subset. Let x iIUi.Then there exists j I such that x Uj and > 0 such that B(x, ) Uj . ThenB(x, ) Uj iIUi.Remark 31 (Off-syllabus) We saw in Theorem 27 that knowledge of the open sets is sufficientto determine which functions are continuous. So the open sets are more fundamental to conti-nuitythan the actual metric (many different metrics will lead to the same family of open sets).In line with this thinking a topological space is defined as follows.A topological space (X, T ) is a set X together with a family T of subsets of X with thefollowing properties:(i) T and X T ;(ii) T is closed under finite intersections;(iii) T is closed under arbitrary unions.T is called the topology and U is said to be open when U T .So all metric spaces are topological spaces though not all topologies are induced by metrics.However a topological space (which is a much more general notion) is sufficient to describe manyaspects of continuity and indeed the proofs are often neater and more natural in that contextwithout the unnecessary and messy inequalities associated with a metric. Those interested inthis should continue with the Hilary Term long option in Topology.14 OPEN AND CLOSED SETS 3. Example 32 An infinite collection of open sets need not have an open intersection. For ex-ample,consider Ui = (i, i) for i > 0 which have intersection {0} which is not open.Definition 33 C M is said to be closed in M (or simply closed) if MC is open.Remark 34 Note that open and closed are not opposites of one another! Sets may be open,closed, neither or both, as we shall see.Example 35 (a) and M are both open and closed in M. That is open is "vacuously true"and M is clearly open by definition.(b) The interval [0, 1) is neither open nor closed in R. To see this, note it is not open as 0is not an interior point of [0, 1) and that the set is not closed as 1 is not an interior point ofthe complement.(c) The subset [0,) (0,) R2 is neither open nor closed. The point (0, 1) is notinterior to the subset and (1, 0) is not interior to its complement.(d) The subset Z2 is closed but not open in R2.(e) The set {(x, y) : x2 + xy + y2 < 1} is open. To see this we can note that f : R2 Rgiven by f(x, y) = x2+xy+y2 is continuous and then the set in question is f1((, 1)) whichis the preimage by a continuous function of an open set (Theorem 27).(f) Let M = [0, 1] [2, 3]. Then [0, 1] is both open and closed in M. Likewise [2, 3] . (Wewill comment more on such examples in the chapter Subspaces.)Proposition 36 Let x M and > 0. Then B(x, ) is open and B(x, ) is closed.Proof. Firstly if y B(x, ) then let = 12 ( d(y, x)) > 0. By the triangle inequality, ifz B(y, ) thend(z, x) d(z, y) + d(y, x) < + d(y, x) = + d(y, x)2< + 2= .Hence B(y, ) B(x, ) and B(x, ) is open.To prove the second result, take y MB(x, ), Then d(x, y) > and set = 12(d(x, y)) >0. If d(z, y) < then by the triangle inequalityd(z, x) d(x, y) d(z, y) > d(x, y) = + > and hence z MB(x, ). That is MB(x, ) is open and B(x, ) is closed.Proposition 37 Singleton sets are closed.Proof. Let x M. If y= x we may set = d(x, y) > 0. Then x / B(y, ) and hence y isinterior to M{x}.Example 38 Note, when M = (0, 1) {2}, that BM(2, 1) = BM(2, 1) = {2}. In this case thisopen ball is also closed.OPEN AND CLOSED SETS 15 4. Definition 39 Let S M and x M. We say that x is a limit point of S or an accumu-lationpoint of S if for any > 0(B(x, ) S) {x}= .i.e. there are points of S arbitrarily close to x other than x, which may or may not be in S.We will denote the set of limit points of S as S.Limit Point in setInterior PointLimit Point not in setNot aLimitPointInterior and limit points of a setExample 40 (a) Let M = R and S = [0, 1). Then S = [0, 1] . But if M = S = [0, 1), thenS = [0, 1).(b) Let M = R and S = (, 1) {2}. Then S = (, 1]. So a point of S need not be apoint of S.Proposition 41 C M is closed if and only if it contains all its limit points.Proof. Note that x is an interior point of MC if and only if x is not a limit point of C. SoC is closed MC is open every point of MC is an interior point no point of MC is a limit point of C C contains all its limit points.The results corresponding to Proposition 30 for closed sets are:Proposition 42 (a) The union of finitely many closed sets is closed.(b) An arbitrary intersection of closed sets is closed.Proof. These simply follow from applying De Morgans laws to the equivalent properties ofopen subsets shown in Proposition 30.And we also have:16 OPEN AND CLOSED SETS 5. Corollary 43 (To Theorem 27) Let f : M N be a map between metric spaces M and N.Then f is continuous if and only if whenever C is closed in N then f1(C) is closed in M.Proof. Noting that for any S N we have f1 (NS) = Mf1 (S) , we havef is continuous f1 (U) is open in M whenever U is open in N f1 (Y C) is open in M whenever C is closed in N Mf1 (C) is open in M whenever C is closed in N f1 (C) is closed in M whenever C is closed in N.Corollary 44 Open balls are open and closed balls are closed.Proof. We showed this in Proposition 36. However this is now an easy application of Theorem27 and Corollary 43 to the continuous function f(x) = d(x, a) from Proposition 23 asB(a, ) = f1((, )) and B(a, ) = f1((, ]).Example 45 The sphere x2 +y2 +z2 = a2 is closed in R3. This is because {a2} is closed in Rand f : R3 R is continuous wheref(x, y, z) = x2 + y2 + z2.The sphere is then f1({a2}).Example 46 Let c denote the subset of l consisting of all convergent sequences. Then c is aclosed subset of l.Solution. Say that x(k)n is a sequence in c which converges to (Xn) in l as k . Weaim to show that (Xn) is in c; we will do this by showing (Xn) is Cauchy and so convergent.Let > 0. As x(k)n (Xn) in l thenn (Xn)x(k)n x(k)= supn Xn 0 as k .So there exists K such thatx(k)n Xn< /3 for k K and all n.As x(K)n is convergent then it is Cauchy. So there exists N such that x(K)n x(K)m < /3 for n,m N.Thus for m, n N we have|Xm Xn| Xm x(K)m + x(K)m x(K)n + x(K)n Xn< /3 + /3 + /3 = .OPEN AND CLOSED SETS 17 6. Example 47 Given a metric space M, x M and > 0, the limit points of B(x, ) need notinclude B(x, ).Solution. Take the discrete metric on a set M and x M. Then B(x, 1) = {x} whilstB(x, 1) = M. However given any y= x we see that x / B(y, 1) and so y is not a limit point ofB(x, 1). In fact there are no limit points of B(x, 1).18 OPEN AND CLOSED SETS